32.3: Neutron Activation Methods

Few analytes are naturally radioactive. For many analytes, however, we can induce radioactivity by irradiating the sample with neutrons in a process called neutron activation analysis (NAA). The radioactive element formed by neutron activation decays to a stable isotope by emitting a gamma ray, and, possibly, other nuclear particles. The rate of gamma-ray emission is proportional to the analyte’s initial concentration in the sample. For example, if we place a sample containing non-radioactive $_{13}^{27}\text{Al}$ in a nuclear reactor and irradiate it with neutrons, the following nuclear reaction takes place.

$$_{13}^{27} \mathrm{Al}+_{0}^{1} \mathrm{n} \longrightarrow_{13}^{28} \mathrm{Al} \nonumber$$

The radioactive isotope of ²⁸Al has a characteristic decay process that includes the release of a beta particle and a gamma ray.

$$_{13}^{28} \mathrm{Al} \longrightarrow_{14}^{28} \mathrm{Al} + _{-1}^{0} \beta + \gamma \nonumber$$

When irradiation is complete, we remove the sample from the nuclear reactor, allow any short-lived radioactive interferences to decay into the background, and measure the rate of gamma-ray emission.

The initial activity at the end of irradiation depends on the number of atoms that are present. This, in turn, is a equal to the difference between the rate of formation for $_{13}^{28}\text{Al}$ and its rate of disintegration

$$\frac {dN_{_{13}^{28} \text{Al}}} {dt} = \Phi \sigma N_{_{13}^{27} \text{Al}} - \lambda N_{_{13}^{28} \text{Al}} \label{13.5}$$

where $\Phi$ is the neutron flux and $\sigma$ is the reaction cross-section, or probability that a $_{13}^{27}\text{Al}$ nucleus captures a neutron. Integrating Equation $\ref{13.5}$ over the time of irradiation, t_i, and multiplying by $\lambda$ gives the initial activity, A₀, at the end of irradiation as

$$A_0 = \lambda N_{_{13}^{28}\text{Al}} = \Phi \sigma N_{_{13}^{27}\text{Al}} (1-e^{-kt}) \nonumber$$

If we know the values for A₀, $\Phi$, $\sigma$, $\lambda$, and t_i, then we can calculate the number of atoms of $_{13}^{27}\text{Al}$ initially present in the sample.

A simpler approach is to use one or more external standards. Letting $(A_0)_x$ and $(A_0)_s$ represent the analyte’s initial activity in an unknown and in an external standard, and letting $w_x$ and $w_s$ represent the analyte’s weight in the unknown and in the external standard, we obtain the following pair of equations

$$\left(A_{0}\right)_{x}=k w_{x} \label{13.6}$$

$$\left(A_{0}\right)_{s}=k w_{s} \label{13.7}$$

that we can solve to determine the analyte’s mass in the sample.

As noted earlier, gamma ray emission is measured following a period during which we allow short-lived interferents to decay into the background. As shown in Figure 32.3.1 , we determine the sample’s or the standard’s initial activity by extrapolating a curve of activity versus time back to t = 0. Alternatively, if we irradiate the sample and the standard at the same time, and if we measure their activities at the same time, then we can substitute these activities for (A₀)_x and (A₀)_s. This is the strategy used in the following example.

Figure 32.3.1 . Plot of gamma-ray emission as a function of time showing how the analyte’s initial activity is determined.

Example 32.3.1

The concentration of Mn in steel is determined by a neutron activation analysis using the method of external standards. A 1.000-g sample of an unknown steel sample and a 0.950-g sample of a standard steel known to contain 0.463% w/w Mn are irradiated with neutrons for 10 h in a nuclear reactor. After a 40-min delay the gamma ray emission is 2542 cpm (counts per minute) for the unknown and 1984 cpm for the external standard. What is the %w/w Mn in the unknown steel sample?

Solution

Combining equations $\ref{13.6}$ and $\ref{13.7}$ gives

$$w_{x}=\frac{A_{x}}{A_{s}} \times w_{s} \nonumber$$

The weight of Mn in the external standard is

$$w_{s}=\frac{0.00463 \text{ g } \text{Mn}}{\text{ g } \text { steel }} \times 0.950 \text{ g} \text { steel }=0.00440 \text{ g} \text{ Mn} \nonumber$$

Substituting into the above equation gives

$$w_{x}=\frac{2542 \text{ cpm}}{1984 \text{ cpm}} \times 0.00440 \text{ g} \text{ Mn}=0.00564 \text{ g} \text{ Mn} \nonumber$$

Because the original mass of steel is 1.000 g, the %w/w Mn is 0.564%.

Among the advantages of neutron activation are its applicability to almost all elements in the periodic table and that it is nondestructive to the sample. Consequently, NAA is an important technique for analyzing archeological and forensic samples, as well as works of art.